Differentiable function with no second derivative at $0$?
0
$begingroup$
What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?
calculus analysis examples-counterexamples
edited Dec 6 '18 at 16:59
Andrews
3901317
asked Dec 6 '18 at 16:49
Y.Z.Y.Z.
173
$endgroup$
add a comment |
0
$begingroup$
What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?
calculus analysis examples-counterexamples
edited Dec 6 '18 at 16:59
Andrews
3901317
asked Dec 6 '18 at 16:49
Y.Z.Y.Z.
173
$endgroup$
3
$begingroup$
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
$endgroup$
– Michael Hoppe
Dec 6 '18 at 16:57
$begingroup$
@Eric Maybe only integral of absolute value?
$endgroup$
– Юрій Ярош
Dec 6 '18 at 17:01
$begingroup$
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
$endgroup$
– Eric
Dec 6 '18 at 17:04
add a comment |
0
0
0
$begingroup$
What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?
calculus analysis examples-counterexamples
edited Dec 6 '18 at 16:59
Andrews
3901317
asked Dec 6 '18 at 16:49
Y.Z.Y.Z.
173
$endgroup$
What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?
calculus analysis examples-counterexamples
calculus analysis examples-counterexamples
edited Dec 6 '18 at 16:59
Andrews
3901317
asked Dec 6 '18 at 16:49
Y.Z.Y.Z.
173
edited Dec 6 '18 at 16:59
Andrews
3901317
asked Dec 6 '18 at 16:49
Y.Z.Y.Z.
173
edited Dec 6 '18 at 16:59
Andrews
3901317
edited Dec 6 '18 at 16:59
Andrews
3901317
edited Dec 6 '18 at 16:59
Andrews
3901317
3901317
asked Dec 6 '18 at 16:49
Y.Z.Y.Z.
173
asked Dec 6 '18 at 16:49
Y.Z.Y.Z.
173
asked Dec 6 '18 at 16:49
Y.Z.Y.Z.
173
173
3
$begingroup$
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
$endgroup$
– Michael Hoppe
Dec 6 '18 at 16:57
$begingroup$
@Eric Maybe only integral of absolute value?
$endgroup$
– Юрій Ярош
Dec 6 '18 at 17:01
$begingroup$
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
$endgroup$
– Eric
Dec 6 '18 at 17:04
add a comment |
3
$begingroup$
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
$endgroup$
– Michael Hoppe
Dec 6 '18 at 16:57
$begingroup$
@Eric Maybe only integral of absolute value?
$endgroup$
– Юрій Ярош
Dec 6 '18 at 17:01
$begingroup$
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
$endgroup$
– Eric
Dec 6 '18 at 17:04
3
3
$begingroup$
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
$endgroup$
– Michael Hoppe
Dec 6 '18 at 16:57
$begingroup$
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
$endgroup$
– Michael Hoppe
Dec 6 '18 at 16:57
$begingroup$
@Eric Maybe only integral of absolute value?
$endgroup$
– Юрій Ярош
Dec 6 '18 at 17:01
$begingroup$
@Eric Maybe only integral of absolute value?
$endgroup$
– Юрій Ярош
Dec 6 '18 at 17:01
$begingroup$
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
$endgroup$
– Eric
Dec 6 '18 at 17:04
$begingroup$
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
$endgroup$
– Eric
Dec 6 '18 at 17:04
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 '18 at 16:59
AndreiAndrei
11.8k21026
$endgroup$
$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32
$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028751%2fdifferentiable-function-with-no-second-derivative-at-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 '18 at 16:59
AndreiAndrei
11.8k21026
$endgroup$
$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32
$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22
add a comment |
1
$begingroup$
So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 '18 at 16:59
AndreiAndrei
11.8k21026
$endgroup$
$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32
$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22
add a comment |
1
1
1
$begingroup$
So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 '18 at 16:59
AndreiAndrei
11.8k21026
$endgroup$
So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 '18 at 16:59
AndreiAndrei
11.8k21026
answered Dec 6 '18 at 16:59
AndreiAndrei
11.8k21026
answered Dec 6 '18 at 16:59
AndreiAndrei
11.8k21026
answered Dec 6 '18 at 16:59
AndreiAndrei
11.8k21026
11.8k21026
$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32
$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22
add a comment |
$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32
$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22
$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32
$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32
$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22
$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028751%2fdifferentiable-function-with-no-second-derivative-at-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
$endgroup$
– Michael Hoppe
Dec 6 '18 at 16:57
$begingroup$
@Eric Maybe only integral of absolute value?
$endgroup$
– Юрій Ярош
Dec 6 '18 at 17:01
$begingroup$
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
$endgroup$
– Eric
Dec 6 '18 at 17:04